Question

50 mL of 0.060 M K₂CrO₄ is mixed with 50 mL of 0.080 M AgNO₃. Calculate the following:

a. The solubility of Ag₂CrO₄ (K_sp = 1.9 X 10^-12) in the solution in moles per liter.

b. The concentrations of the following ions Ag⁺, CrO₄^-2, K⁺, and NO₃^-.

Answer 1

50 mL of 0.060 M K₂CrO₄ is mixed with 50 mL of 0.080 M AgNO₃.

a. Ksp of Ag2CrO4 = 1.9 x 10^-12 = [Ag+]^2[CrO4^2-]

The ratio of Ag2CrO4 to Ag+ is 1:2, and ratio to CrO4- is 1:1

moles of K2CrO4 = M x L = 0.06 x 0.05 = 0.003 mols

moles of AgNO3 = 0.08 x 0.05 = 0.004 mols

We know,

K2CrO4 + 2AgNO3 ----> Ag2CrO4 + 2KNO3

So, 1 mole of K2CrO4 reacts with 2 moles of AgNO3 to give 1 mole of Ag2CrO4 and 2 moles of KNO3

thus, 0.004 mols of AgNO3 will react with 0.002 mols of K2CrO4 to give 0.002 moles of Ag2CrO4

remaning K2CrO4 = 0.003 - 0.002 = 0.001 mols

total volume of solution = 0.05 + 0.05 = 0.1 L

Molar concentration of Ag2CrO4 in 0.1 L = 0.002/0.1 = 0.02 M

Ag2CrO4 will solubilize and form ions, lets say x moles has ionized,

So, Ksp for Ag2CrO4 = [2x]^2[x]

1.9 x 10^-12 = 4x^3

x = 0.000078 M is the molar solubility of Ag2CrO4

b. concentration of ions in solution,

molar concentration of CrO42- = 0.001/0.1 = 0.01 M

CrO4^2- = 0.000078 M
Ag+ = 2x0.000078 = 0.000156 M
K+ = 0.000078/2 = 0.000039 M

NO3^-1 = 0.02 M